I am wondering why for a Brownian Motion $B_t$ and where $s < t$:
$$ E(sin(B_{t-s})) = 0 $$ ?
I can't understand why, can anyone see?
2026-04-05 18:23:53.1775413433
Why is $E(sin(B_{t-s})) = 0$?
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$\sin$ is an odd, bounded, function. If the distribution of $X$ is symmetric about zero, then $E[\sin(X)]=0$.